# 567.Permutation-in-String

## 题目描述

Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
Example 1:
Input: s1 = "ab" s2 = "eidbaooo"
Output: True
Explanation: s2 contains one permutation of s1 ("ba").
Example 2:
Input:s1= "ab" s2 = "eidboaoo"
Output: False
Note:
The input strings only contain lower case letters.
The length of both given strings is in range [1, 10,000].

## 代码

### Approach #1 Using Array Map

class Solution {
public boolean checkInclusion(String s1, String s2) {
if (s1.length() > s2.length()) return false;
int[] s1map = new int[26];
for (int i = 0; i < s1.length(); i++)
s1map[s1.charAt(i) - 'a']++;
for (int i = 0; i < s2.length() - s1.length(); i++) {
int[] s2map = new int[26];
for (int j = 0; j < s1.length(); j++) {
s2map[s2.charAt(i + j) - 'a']++;
}
if (matches(s1map, s2map)) {
return true;
}
}
return false;
}
private boolean matches(int[] s1map, int[] s2map) {
for (int i = 0; i < 26; i++) {
if (s1map[i] != s2map[i]) {
return false;
}
}
return true;
}
}

### Approach #2 Sliding Windwow

public class Solution {
public boolean checkInclusion(String s1, String s2) {
if (s1.length() > s2.length())
return false;
int[] s1map = new int[26];
int[] s2map = new int[26];
for (int i = 0; i < s1.length(); i++) {
s1map[s1.charAt(i) - 'a']++;
s2map[s2.charAt(i) - 'a']++;
}
for (int i = 0; i < s2.length() - s1.length(); i++) {
if (matches(s1map, s2map)) return true;
s2map[s2.charAt(i + s1.length()) - 'a']++;
s2map[s2.charAt(i) - 'a']--;
}
return matches(s1map, s2map);
}
public boolean matches(int[] s1map, int[] s2map) {
for (int i = 0; i < 26; i++) {
if (s1map[i] != s2map[i])
return false;
}
return true;
}
}

### Approach #3 Optimized Sliding Window

public class Solution {
public boolean checkInclusion(String s1, String s2) {
if (s1.length() > s2.length())
return false;
int[] s1map = new int[26];
int[] s2map = new int[26];
for (int i = 0; i < s1.length(); i++) {
s1map[s1.charAt(i) - 'a']++;
s2map[s2.charAt(i) - 'a']++;
}
int count = 0;
for (int i = 0; i < 26; i++)
if (s1map[i] == s2map[i])
count++;
for (int i = 0; i < s2.length() - s1.length(); i++) {
int r = s2.charAt(i + s1.length()) - 'a', l = s2.charAt(i) - 'a';
if (count == 26)
return true;
s2map[r]++;
if (s2map[r] == s1map[r])
count++;
else if (s2map[r] == s1map[r] + 1)
count--;
s2map[l]--;
if (s2map[l] == s1map[l])
count++;
else if (s2map[l] == s1map[l] - 1)
count--;
}
return count == 26;
}
}